hmwk4sol_a

hmwk4sol_a - Math 202a Michael Phillips 9/26/07 Problem 1....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 202a Michael Phillips 9/26/07 Problem 1. (a.) Pick any x ∈ [0 , 2 π ]. sin kx has period 2 π k , for any δ > 0, we can take k large enough so that 2 π k < δ- i.e. so that at least two periods of sin kx wholly fit within B ( x,δ ). So, we would then have sin kx = sin k ( x ± 2 π k ) with 2 π k < δ . But then we could choose an x 1 ∈ [ x,x + 2 π k ] such that | sin kx- sin kx 1 | ≥ 1. Hence, we equicontinuity fails since we cannot guarantee the conditions for < 1. (b.) Consider the fixed point x = 1 of all the { x n } . For 0 < x < 1, we know that lim n →∞ = 0. Hence, for any δ > 0, we can find an N so that if n > N , we have that (1- δ ) n < for any > 0. So, we can find an N so that if n > N , we have that | (1) n- (1- δ ) n | is within of one. Hence the conditions for equicontinuity fail at x = 1 if we ask for any < 1. We use Problem 3. We have that the derivatives of the family, i.e. nx n- 1 , are uniformly bounded by 1 in [0 , 1 / 2]. Why? They are all bounded above by 1, the derivative of the constant function at x = 1 / 2. Hence, since each derivative monotonically decreases towards zero, we have that they all stay below 1. Problem 2. Take any 1 > 0. A ( f )( x 1 )- A ( f )( x 2 ) = φ ( x 1 )- φ ( x 2 )+ R b a ( K ( x 1 ,y )- K ( x 2 ,y )) f ( y ) dy . Now since φ is a continuous function on compact domain, we know that it is uniformly continuous on that domain. Likewise, for fixed y , we have that K ( x,y ) is a continuous function of x on a compact domain, hence, is uniformly continuous. So, we have ∀ > 0, a pair δ,δ 1 > 0 so that ∀ x 1 ,x 2 ∈ [ a,b ] we have | ( x 1 ,y )- ( x 2 ,y ) | < δ ⇒ | φ ( x 1 )- φ ( x 2 ) | < and so that | x 1- x 2 | < δ 1 ⇒ | K ( x 1 ,y )- K ( x 2 ,y ) | < . Choose the minimum of these two, call it δ . Now, | x 1- x 2 | < δ , we have | A ( f )( x 1 )- A ( f )( x 2 ) | ≤ + M | a- b | = (1 + M | a- b | ), where M is the bound over all functions in our set of functions in C ([ a,b ]) that is guaranteed by assumption. We want this number to be less than our arbitrary 1 , so we choose δ,δ 1 so that < 1 M | a- b | +1 , then, taking the minimum of δ,δ 1 gives us what we need. Problem 3. Take any > 0. Let δ = /M , where M is such that ∀ f ∈ E and ∀ x ∈ [ a,b ] we have that | f ( x ) | ≤ M . So, suppose | x 1- x 2 | < δ for any x 1 ,x 2 ∈ [ a,b ]. Then | f ( x 1 )- f ( x 2 ) | = | R x 2 x 1 f ( y ) dy | ≤ R x 2 x 1 | f ( y ) | dy ≤ M | x 1- x 2 | ≤ Mδ < , as needed....
View Full Document

This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

Page1 / 4

hmwk4sol_a - Math 202a Michael Phillips 9/26/07 Problem 1....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online